While the decibel scale can be used in many disciplines, we shall restrict our attention to its use in acoustics, specifically its use in measuring the intensity level of sound. The Sound Intensity Level (measured in decibels) of a sound intensity (measured in watts per square meter) is given by Compute Damage to your hearing can start with short term exposure to sound levels around 115 decibels. What intensity is needed to produce this level

The farmer uses this technique to protect the produce from freezing because as the water in the tubs begins to freeze, it releases heat. This heat can help to keep the surrounding air in the shed above freezing, protecting the produce from freezing as well. Additionally, as the water in the tubs freezes, it releases latent heat of fusion, which is the energy required to change a substance from a solid to a liquid. This energy is also released in the form of heat, further helping to keep the produce from freezing.

Answer:

Volume measures liquid

Explanation:

Volume can be measured in meters and centimeters

Answer:

The force between two magnetic poles will increase as their separation is increased.

Answer:

Anywhere above 12000N in order to decelerate the train since F=mass*acceleration

Answer:

Atoms in a solid are close and tightly packed.They vibrate about fixed positions .They are arranged in a regular LATTICE.They cannot be compressed nor can they change shape.They have strong forces of attraction between them so they cannot move in order to move they have to gain ENERGY. Whereas the atoms in a liquid are not that close compared to the ones in solid.They take the shape of the container but their volume remains fixed.They have weak forces of attraction between atoms and have more energy which enables them to move.

Explanation:

It would take approximately 546 days for the decay rate of the sample of this radioactive isotope to fall to 0.58 of its initial rate.

1. The decay rate of a radioactive isotope is proportional to the number of radioactive atoms present in the sample at any given time.

2. The decay rate can be expressed as a function of time using the formula: R(t) = R₀ * \(e^{(-\lambda t\)), where R(t) is the decay rate at time t, R₀ is the initial decay rate, λ is the decay constant, and e is the base of the natural logarithm.

3. The half-life of a radioactive isotope is the time it takes for half of the radioactive atoms in a sample to decay. In this case, the half-life is given as 210 days.

4. Using the half-life, we can find the decay constant (λ) using the formula: λ = ln(2) / T₁/₂, where ln(2) is the natural logarithm of 2 and T₁/₂ is the half-life.

5. Substituting the given half-life into the formula, we have: λ = ln(2) / 210.

6. Now, we need to find the time it takes for the decay rate to fall to 0.58 of its initial rate. Let's call this time "t".

7. Using the formula for the decay rate, we can write: 0.58 * R₀ = R₀ * e^(-λt).

8. Simplifying the equation, we get: 0.58 = \(e^{(-\lambda t\)).

9. Taking the natural logarithm of both sides, we have: ln(0.58) = -λt.

10. Substituting the value of λ from step 5, we get: ln(0.58) = -(ln(2) / 210) * t.

11. Solving for t, we have: t = (ln(0.58) * 210) / ln(2).

12. Evaluating the expression, we find: t ≈ 546.

13. Therefore, it would take approximately 546 days for the decay rate of the sample of this radioactive isotope to fall to 0.58 of its initial rate.

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Answer:

Explanation:

The mass of a substance when given the density and volume can be found by using the formula

mass = Density × volume

From the question we have

mass = 1.5 × 8

We have the final answer as

Hope this helps you

Newton's second law allows to find the result for the acceleration of the file and the ball is:

Newton's second law says that the force is proportional to the product of the acceleration for the mass of a body.

                F = ma

Where F is the force, the acceleration and m the mass of the body.

                \(a = \frac{F}{m}\)  

In this case we have a baseball with a mass m and a filing cabinet with a mass M.

           M> m

Therefore, using Newton's second law, the acceleration of the cabinet must be less than the acceleration of the ball since the mass of the cabinet is greater than mass ball.

In conclusion using Newton's second law we can find the result for the acceleration of the filing cabinet and the ball is:

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Answer:

The Velocity is 667 N Force Hope it helps<333 and the cat is for your encuragement..

The final velocity of the football when it is pushed to a distance of 0.2 m is 62.4 m/s.

Kinetic friction is defined as the resistive force exerted by a surface on an object whose surface is in contact with it. It opposes the relative motion between the surfaces in contact.

Here,

Mass  of  the football,  m = 0.05 kg

Force acting on it, F = 8.3 N

Frictional force acting on the ball F(k) = 0.5 N

If an energy diagram is drawn,

The football is initially at rest. When a constant force of 8.3 N is exerted it starts to get pushed, so that it moves. The force of friction exerted on the moving football brings it to a steady motion with constant speed.

Net force acting on the football, Fnet = F - F(k)

   ma = 8.3 - 0.5 = 7.8 N

So, Acceleration, a = Fnet/m

  a = 7.8/0.05

  a = 156 m/s²

Using the equation of motion,

v² = u² + 2as

v² = 0 + 2x 156x 0.2

v² = 62.4 m/s

Hence,

The final velocity of the football when it is pushed to a distance of 0.2 m is 62.4 m/s.

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Attaching the image here.

Gerard's average velocity for the ride is 6 meters per second.

To find Gerard's average velocity for the ride, we can use the formula:

Average velocity = Total displacement / Total time

First, we need to convert the distance traveled from kilometers to meters:

7.20 kilometers * 1000 = 7200 meters

Next, we convert the time from minutes to seconds:

20.0 minutes * 60 = 1200 seconds

Now, we can calculate the total displacement by subtracting the initial position from the final position. Since Gerard is riding directly east, there is no change in the east-west direction, so the displacement is equal to the distance traveled:

Total displacement = 7200 meters

Finally, we substitute the values into the average velocity formula:

Average velocity = 7200 meters / 1200 seconds

Average velocity = 6 meters per second

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Answer:

i believe the answer is B only because i looked it up on g00gle so take what i say with a GRAIN of salt all puns intended

Explanation:

Answer:

138

Explanation:

(since there's a blank next to "9x y j", I'm assuming the y of F(x, y) is 9x\(\sqrt{y}\) j)

1) find the partial derivative of each:

\((6y^{\frac{3}{2} })i + (9x\sqrt{y} )j\)  

 \(f_{x} =(6y^{\frac{3}{2} })_{x} = \int\limits{6y^{\frac{3}{2} }} \, dx = 6xy^{\frac{3}{2} } +c \\\\f_{y} = (9x\sqrt{y} )_{y} = \int\limits{9xy^{\frac{1}{2} } } \, dy = 9x(\frac{2}{3} )y^{\frac{3}{2} } +c = 6xy^{\frac{3}{2} } +c\)

2) use partial integrals to make gradient of f:

take whatever you got from partial integral and add them together (if they repeat, just use it once)

\(F =\) Vf (V = gradient of)

\(F(x, y) = 6xy^{\frac{3}{2} }\)

3) Evaluate the integrals with given points:

Integral of F dotted with dr = F(point B) - F(point A) = F(3, 4) - F(1, 1)

\(F(point B) = 6(3)(4)^{\frac{3}{2} }\\F(Point A) = 6(1)(1)^{\frac{3}{2}}\\F(point B) - F(pointA) = 6(3)(4)^{\frac{3}{2} }-(6(1)(1)^{\frac{3}{2}})\)

= 144 - 6 = 138 units of work

Work done by the force field F in moving an object from A to B = 138 J

Given data :

Force field F(x,y) = \(6y^{\frac{3}{2} }i + (9x\sqrt{y} ) j\)  

Step 1 : determine the partial derivatives of the vector quantity

Fx = ∫ \(6y^{\frac{3}{2} }i = 6xy^{\frac{3}{2} } + c\)

Fy = ∫ \((9x \sqrt{y})_{y} = 9x(\frac{2}{3})y^{\frac{3}{2} } + c\)

Equating the partial derivatives :  

\(9x(\frac{2}{3})y^{\frac{3}{2} } + c\)  = \(6xy^{\frac{3}{2} } + c\)

therefore the gradient of F  i.e. F = vF  = F( x,y ) = \(6xy^{\frac{3}{2} }\)

Next step : Determine the work done

Work done ( F.dr ) = [ F(point b ) = F( 3,4 ) ]  - [ F(point A) = F( 1,1 ) ]

F(3,4 ) = 6(3)(4)\(^{\frac{3}{2} }\)  = 144

F( 1,1 )  = 6(1)(1)\(^{\frac{3}{2} }\)    = 6

Therefore the work done by the force field = 144 - 6 = 138 J

Hence we can conclude that the work done by the force field F is = 138 J

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A procedure known as magnetic separation can be used to extract magnetic elements from coal.

This method makes use of the magnetic characteristics of some materials to distinguish them from non-magnetic materials like coal. A description of how magnetic separation can be used to remove magnetic components from coal is given below:

Putting a magnetic field around the coal and magnetic material mixture is the first step in the magnetization process. This can be achieved by creating an electromagnetic field or by putting the mixture close to a powerful magnet.

Magnetism: Magnetic materials, such as iron atoms or magnetite that are frequently found in coal, will be drawn to the magnetic field and become magnetized. They line up their magnetic moments with the magnetic field's direction.

Separation: The magnetic coal components can be physically separated from the non-magnetic coal once they have been magnetized. To create this separation, there are numerous techniques:

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Explanation:

density, mass of a unit volume of a material substance. The formula for density is d = M/V, where d is density, M is mass, and V is volume. Density is commonly expressed in units of grams per cubic centimetre.

The relative density of a substance is defined as the ratio of the density of that substance to the density of water at 4oC. It is also defined as the ratio of the mass of substance to the mass of equal volume of water at 4oC. i.e., R.D. = Mass of the substance / mass of an equal volume of water at 4oC

The net force on particle q₂, located between particles q₁ and q₃, is approximately 189000 N. The force exerted by particle q₁ on q₂ is positive and equals 252000 N, while the force exerted by particle q₃ on q₂ is negative and equals -63000 N.

To find the net force on particle q₂, we need to calculate the individual forces exerted on q₂ by particles q₁ and q₃ and then determine their sum.

The force between two charged particles can be calculated using Coulomb's law:

F = k * |q₁ * q₂| / r²

Where F is the force between the particles, k is the electrostatic constant (k ≈ 9.0 x \(10^9\) Nm²/C²), q₁ and q₂ are the charges of the particles, and r is the distance between them.

First, let's calculate the force exerted on q₂ by q₁:

F₁₂ = k * |q₁ * q₂| / r₁₂²

F₁₂ = (9.0 x \(10^9\) Nm²/C²) * |(8.0 μC) * (3.5 μC)| / (0.10 m)²

F₁₂ ≈ 252000 N

The force is positive because q₁ and q₂ have opposite charges.

Next, let's calculate the force exerted on q₂ by q₃:

F₂₃ = k * |q₂ * q₃| / r₂₃²

F₂₃ = (9.0 x \(10^9\)Nm²/C²) * |(3.5 μC) * (-2.5 μC)| / (0.15 m)²

F₂₃ ≈ -63000 N

The force is negative because q₂ and q₃ have the same charge.

Finally, we can find the net force on q₂ by summing the individual forces:

Net force = F₁₂ + F₂₃

Net force = 252000 N + (-63000 N)

Net force ≈ 189000 N

The net force on particle q₂ is approximately 189000 N.

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Answer:

Do not see a picture or graph but suspect it would show the golf ball falling faster and striking the ground slightly before the soccer ball.

Probably D:  Soccer ball was affected by air resistance more than the golf ball.

Explanation:

Even though heavier, friction loss of the greater surface area soccer ball will counter pull of gravity more than the compact golf ball.

In a vacuum, (no friction) both objects fall at the same rate regardless of mass.

Answer:

0.838

Explanation:

The ratio v/c of the speed v to the speed c of light in a vacuum is shown below:

Given that

\(\triangle t_0 = 24\ seconds\) = time interval for one revolution

\(\triangle t = 44\ seconds\) = time interval measured with speed v

based on the given information, the ratio v/c  of the speed v to the speed c of light in a vacuum is

\(\triangle t = \frac{\triangle t_0}{\sqrt{1 - \frac{v^2}{c^2}}}\)

\({\sqrt{1 - \frac{v^2}{c^2}}} = \frac{\triangle t_0}{\triangle t}\)

Now squaring both the sides

\(\frac{v^2}{c^2} = 1 - \frac{(\triangle t_0)^2}{(\triangle t)^2}\)

Now remove the squaring root from both the sides and putting the values

\(\frac{v}{c} = {\sqrt{1 - \frac{(\triangle t_0)^2}{(\triangle t)^2}\)

\(= {\sqrt{1 - \frac{(24)^2}{(44)^2}\)

= 0.838

The evidence that Wegener did NOT use to support his idea of continental drift is "the thickness of layers of ice in the Antarctic.

option C

Wegener's primary evidence for continental drift included the fit of the coastlines of different continents, the distribution of fossils across different continents, and the alignment of rock strata on different continents.

So the thickness of layers of ice in the Antarctic, was not used by Wegener to support his idea of continental drift. While this evidence is important for supporting the theory of glaciation, it is not relevant to the theory of continental drift.

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Answer:

14.57 ohms

Explanation:

Here in the figure ,Rb & R₄are in series  & also  Rc & R₅ are in series. As they are in series , ( Rb + R₄ ) & (Rc & R₅) are in parallel . So the equivalent resistance in that branch = ( 2 + 18 ) ║ ( 3 + 12 )

                                          = 20 ║ 15

                                          = (20×15) / (20 + 15)

                                          = 8.57 ohms

Also Ra ( 6 ohm ) is in series with that branch ,. So the equivalent resistance of the whole circuit = 8.57 + 6 = 14.57 ohms.

total resistance is 26 ohm and voltage across \(R_C\) resister is 1.1554 Ampere.

According to the circuit,

\(R_A\) = 6 Ω.

\(R_B\) = 2 Ω.

\(R_C\) = 3 Ω.

R₄ = 18 Ω.

R₅ = 12 Ω.

V = 10 V.

In the circuit, the series combination of \(R_C\) and R₅ is parallel connected with series combination of \(R_B\) and R₄. And this network is connected with \(R_A\) in series combination.

So, Equivalent resistance of the circuit is; R =  \(R_A\) + ( \(R_C\) + R₅) II (\(R_B\) +R₄)

=   18 Ω + ( 3 Ω +  12 Ω)II(2 Ω+ 18 Ω)

=  18 Ω + 15ΩII20Ω

=   18 Ω  + (15×20)/(15+20)Ω

= 18 Ω + 8.57 Ω

= 26 Ω

So, current flowing through the circuit: I = V/ R = 10V / 26Ω = 0.38 Amp.

Voltage drop across the parallel network; V₁ =  V - I\(R_A\)

= 10V - 0.38×6 V

= 7.72 V.

So, voltage across the resister \(R_C\) = V₁\(R_C\)/\((R_C\) + R₅)

= 7.72 × 3/(3 + 12) Amp

= 1.1544 Amp.

Hence, total resistance is 26 ohm and voltage across specific resister is 1.1554 Ampere.

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If A box of mass 210 kg is pulled from rest with a string of tension 1300n inclined at 35° to the horizontal. if the box moved with a speed of 10m/s and the frictional force between the box and surface is 100 n, Then the distance covered by the box is 10.89 meters.

To calculate the distance covered by the box, we need to analyze the forces acting on it and apply the work-energy principle.

Given:

Mass of the box, m = 210 kg

Tension in the string, T = 1300 N

The angle of inclination, θ = 35°

Frictional force, f = 100 N

Initial speed, u = 0 m/s

Final speed, v = 10 m/s

First, let's resolve the tension force into components parallel and perpendicular to the incline. The parallel component of the tension force can be calculated as:

T_parallel = T * cos(θ)

Next, let's calculate the net force acting on the box along the incline. The net force is given by:

Net force = T_parallel - f

Now, using Newton's second law, we can calculate the acceleration (a) of the box:

Net force = m * a

From the given information, we have the final velocity (v), initial velocity (u), and acceleration (a). We can use the following kinematic equation to calculate the distance covered (s):

v^2 = u^2 + 2as

Rearranging the equation, we get:

s = (v^2 - u^2) / (2a)

Now, let's plug in the given values and calculate the distance covered:

T_parallel = 1300 N * cos(35°) ≈ 1067.35 N

Net force = 1067.35 N - 100 N = 967.35 N

a = (967.35 N) / (210 kg) ≈ 4.61 m/s^2

s = (10 m/s)^2 - (0 m/s)^2 / (2 * 4.61 m/s^2) ≈ 10.89 m

Therefore, the distance covered by the box is approximately 10.89 meters.

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Answer:

Fall

Explanation:

Time period of a wave is the inverse of its frequency. The period of the wave with a frequency of 2.09 Hz is 0.47 seconds.

Frequency of a wave is the number of wave cycles per unit time. Frequency is the inverse of the time period of the wave. Hence, it has the unit of s⁻¹ which is equivalent to Hz.

The higher frequency of a wave indicates more number of wave cycles in a short time. Frequency is directly proportional to the energy and inversely proportional to the  wavelength.

Given the time period of the wave = 2.09 Hz

then frequency = 1/2.09 Hz = 0.47 s.

Therefore, the time period  of the wave is 0.47 seconds.

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Answer:

87

Explanation:

We can measure the flow rate of various liquid by pouring the liquid down an incline and time how long it takes each one to reach the bottom.

option C.

The flow rate of a liquid is how much fluid passes through an area in a particular time.

Flow rate can be articulated in either in terms of velocity and cross-sectional area, or time and volume. As liquids are incompressible, the rate of flow into an area must be equivalent to the rate of flow out of an area.

Generally, the best equipment to measure the flow rate of a liquid is flow meters. In the absence of flow meters, we can other methods such as the one given in the options.

We can pour the various liquid down an incline and time how long it takes each one to reach the bottom.

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Answer:

D.

Explanation

Air pressure.

Answer:air pressure

Explanation:

Answer:

B. Boats need to float.

Answer: drawing of Earth's rock layers

computer design of a space shuttle

3-D representation of a volcano

Explanation:

A scientific model can be defined as the physical, conceptual, and mathematical representation of the facts, processes and events. This helps in demonstration, education, and prediction of patterns of the world based on various aspects of the scientific study.

The drawing of the earth rock layers will help in understanding the generation of rocks.

The computer designed shuttle will help in understanding its function and working in the space.

The 3-D representation of volcano will help in understanding the formation and effect of volcano in the surrounding.

Answer:

1

3

4

Explanation:

a)  The value of its capacitance is 0.90 pF

b)  The charge on the capacitor is 15.3 pC

c)  The magnitude of the uniform electric field between the plates is 6.8 N/C.

given that :

are of the parallel plate capacitor = 2.60 cm²

distance d = 2.5 mm

potential =  17 V

a)  The value of its capacitance is :

C = Eo A / d

  = ( 8.85 × 10⁻¹² C² / N . m ) ( 2.60 × 10⁻⁴ m²) / 2.50 × 10⁻³

  = 0.90 × 10⁻¹² F

 = 0.90 pF

b)  The charge on the capacitor is :

q = CVo

q = 0.90 × 10⁻¹² F × 17 V

q = 15.3 × 10⁻¹² C

q = 15.3 pC

c) The magnitude of the uniform electric field between the plates is :

E = Vo / d

E = 17 / 2.50 × 10⁻³

E = 6.8 N/C

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